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4 years ago

5

Bill is depositing money into a money market account. Let y represent the total amount of money in the account (in dollars). Let

x represente the number of weeks bill has been depositing money into his account. Suppose x and y are related by the equation:
Y= 55x + 1250

A) how much money is bill depositing into the account every week?

B) what was the starting amount of money in the account)?

C) how much money will he have in his account after 26 weeks?

1 answer:

Levart [38]4 years ago

3 0

a) Bill is depositing $55 in the account each week

b) the account started with $1250 in it

c)Bill will have $2680 in his account after 26 weeks

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T or F: If you consume more calories, you must decrease your physical activity to have energy

jarptica [38.1K]

Answer:

F

Step-by-step explanation:

When your losing weight after consuming more calories, you need to increase physical activity to increase the number of calories your body uses for energy.

8 0

3 years ago

Read 2 more answers

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

its 76 degrees fahrenheit at the 6000-foot level of a mountain, and 49 degrees Fahrenheit at the 12000-foot level of the mountai

brilliants [131]

$T = \frac{-9}{2}x + 103$ is the linear equation to find the temperature T at an elevation x on the mountain, where x is in thousands of feet.

<em><u>Solution:</u></em>

The linear equation in slope intercept form is given as:

T = cx + k ------ (i)

Where "t" is the temperature at an elevation x

And x is in thousands of feet

<em><u>Given that its 76 degrees fahrenheit at the 6000-foot level of a mountain</u></em>

Given, when c = 6 thousand ft and $T = 76^{\circ}$ fahrenheit

This implies,

From (i)

76 = c(6) + k

76 = 6c + k

⇒ k = 76 - 6c ----- (ii)

<em><u>Given that 49 degrees Fahrenheit at the 12000-foot level of the mountain</u></em>

Given, when c = 12 thousand ft and $T = 49^{\circ}$ fahrenheit

This implies,

From (i)

49 = c(12) + k

49 = 12c + k

Substitute (ii) in above equation

49 = 12c + (76 - 6c)

49 = 12c + 76 - 6c

49 - 76 = 6c

6c = -27

$c = \frac{-9}{2}$

Substituting the value of c in (ii) we get

$k = 76 - 6( \frac{-9}{2})\\\\k = 76 + 27 = 103$

Substituting the value of c and k in (i)

$T = \frac{-9}{2}x + 103$

Where "x" is in thousands of feet

Thus the required linear equation is found

5 0

3 years ago

The ratio of the geometric mean and arithmetic mean of two numbers is 3:5, find the ratio of the smaller number to the larger nu

IgorC [24]

Answer:

$\frac{1}{9}$

Step-by-step explanation:

Let the numbers be x,y, where x>y

The geometric mean is

$\sqrt{xy}$

The Arithmetic mean is

$\frac{x + y}{2}$

The ratio of the geometric mean and arithmetic mean of two numbers is 3:5.

$\frac{ \sqrt{xy} }{ \frac{x + y}{2} } = \frac{3}{5}$

We can write the equation;

$\sqrt{xy} = 3$

or

$xy = 9 - - - (2)$

l

and

$\frac{x + y}{2} = 5$

or

$x + y = 10 - - - (2)$

Make y the subject in equation 2

$y = 10 - x - - - (3)$

Put equation 3 in 1

$x(10 - x) = 9$

$10x - {x}^{2} = 9$

${x}^{2} - 10x + 9 = 0$

$(x - 9)(x - 1) = 0$

$x =1 \: or \: 9$

When x=1, y=10-1=9

When x=9, y=10-9=1

Therefore x=9, and y=1

The ratio of the smaller number to the larger number is

$\frac{1}{9}$

3 0

3 years ago

45.6/109.2 = x/115<br><br> Is x, 48.02?

MAXImum [283]

Yes correct x is 48.02 here is why.

simplify 45.6/109.2 to 0.417582

multiply both sides by 115

simplify 0.417582 x 115 to 48.021978

switch sides

Answer: x = 48.021978

4 0

3 years ago